What is the vertices (x^2)/8+(y^2)/4 =1 ?

The vertices (x^2)/8+(y^2)/4 =1 is (2sqrt(2),0),(-2sqrt(2),0)

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vertices (x^2)/8+(y^2)/4 =1

vertices

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1

(22, 0), (- 22, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 22, b = 2

(0 + 22, 0), (0 - 22, 0)

Refine

(22, 0), (- 22, 0)

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

(x - 2)^{2} + 4 (y - 3)^{2} = 4

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

\frac{( x - 2 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{25} = 1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{16} = 1

What is the vertices (x^2)/8+(y^2)/4 =1 ?
The vertices (x^2)/8+(y^2)/4 =1 is (2sqrt(2),0),(-2sqrt(2),0)